A primitive cell of a crystal lattice is a set $A$ such that two copies of $A$ which are translated by a lattice vector do not overlap and such that $A$ tiles the entire crystal.
I have read (for example in the german “Festkörperphysik” by Gross, Marx), that all primitive cells have the same size/volume.
Intuitively, this seems plausible, but is there a proof?
My precise, measure theoretic, interpretation of this statement is: If $a_1, \ldots, a_n$ is a basis of $\mathbf{R}^n$ and $A, B \subset \mathbf{R}^n$ are sets such that $(\cup (A+\alpha_1 a_1+\ldots+\alpha_n a_n))C$ and $(\mathbf{R}^n \cup (B+\alpha_1 a_1+\ldots+\alpha_n a_n))^C$ (where the union is over all $\alpha_1, \ldots,\alpha_n ∈ \mathbf{Z}$) are Lebesgue null sets and such that for all $\alpha_1,\ldots,\alpha_n∈\mathbf{Z}$: $(A+\alpha_1 a_1 + \ldots \alpha_n a_n) \cap A$ and $(B+\alpha_1 a_1 + \ldots \alpha_n a_n) \cap B$ are Lebesgue null sets, then $A$ and $B$ have the same Lebesgue measure.